Question

Use tables to evaluate the integrals. You may need to complete the square or change variables to put the integral into a form given in the table. $$ \int \frac{\cos x}{\sin ^{2} x+2 \sin x} d x $$

Short answer

\( \frac{1}{2} \ln \left| \frac{\sin x}{\sin x + 2} \right| + C \)

Step-by-step solution

Simplify the Denominator

To simplify the denominator, consider it in terms of a single trigonometric function. Notice that \( \sin^2 x + 2 \sin x = (\sin x)^2 + 2 \sin x \). Thus, it resembles a quadratic form \( u^2 + 2u \) if we let \( u = \sin x \).

Complete the Square

Complete the square for the expression \( \sin^2 x + 2 \sin x \):\[\sin^2 x + 2 \sin x = (\sin x + 1)^2 - 1\]So the integral becomes:\[\int \frac{\cos x}{(\sin x + 1)^2 - 1} \, dx\]

Use Substitution

Use the substitution \( u = \sin x \), which implies \( du = \cos x \, dx \). Substitute into the integral:\[\int \frac{1}{u^2 + 2u} \, du = \int \frac{1}{(u + 1)^2 - 1} \, du\]

Use Table of Integrals

The integral can be expressed in a form suitable for a standard table of integrals:\[\int \frac{1}{v^2 - a^2} \, du \]where \( v = u+1 \) and \( a^2 = 1 \). The corresponding integral formula is:\[\frac{1}{2a} \ln \left| \frac{v-a}{v+a} \right| + C\]

Apply Formula

Apply the formula with \( v = u + 1 \) and \( a = 1 \):\[\frac{1}{2} \ln \left| \frac{u+1 - 1}{u+1 + 1} \right| + C = \frac{1}{2} \ln \left| \frac{u}{u+2} \right| + C\]

Substitute Back the Original Variable

Recall that \( u = \sin x \). Substitute back to get the final answer in terms of \( x \):\[\frac{1}{2} \ln \left| \frac{\sin x}{\sin x + 2} \right| + C\]

Key concepts

trigonometric substitution

Trigonometric substitution is a powerful technique used in integral calculus to simplify integrals by replacing one variable with a trigonometric function. This method is particularly useful when dealing with integrals containing quadratic expressions. In the given exercise, the substitution involved first simplifying the integral by recognizing a form similar to a quadratic. We used a trigonometric identity to express the function in terms of a single trigonometric function:

• Letting \( u = \sin x \), which simplifies the expression \( \sin^2 x + 2 \sin x \) to \( u^2 + 2u \).
• This kind of relation often calls for a trigonometric substitution to further simplify the integration process.

By these substitutions, we convert the integral into a format that can sometimes allow for a more straightforward evaluation using known integrals or additional techniques such as completing the square.

table of integrals

A table of integrals is essentially a reference tool where common integral forms and their solutions are listed. When faced with a complex integral, like the one in the exercise, sometimes all that's needed is to manipulate the integral into a format that matches one of these standard forms:

• The integral \( \int \frac{1}{v^2 - a^2} \, du \) is a common form found in such tables.
• The solution to this standard integral is given by:\( \frac{1}{2a} \ln \left| \frac{v-a}{v+a} \right| + C \) where \( a \) and \( v \) are adjusted according to the scenario.

Using the table of integrals simplifies the process because once the integral is matched to a form in the table, the solution can be directly applied. This reduces the need for re-calculating or deriving the solution from scratch every time.

completing the square

Completing the square is a method used in algebra to transform a quadratic polynomial into a perfect square trinomial, which can simplify integrals involving quadratic expressions. In the exercise:

• The expression \( \sin^2 x + 2 \sin x \) was transformed to \( (\sin x + 1)^2 - 1 \).
• This substitution helps express the denominator in the form of a difference of squares, making it easier to handle in integral calculus.

Completing the square is crucial because it can turn complicated quadratic expressions into simpler forms that are easier to integrate, often revealing simple matches with standard integral formulas. This step facilitates the use of a table of integrals by bringing the expression closer to a form that can be directly integrated.

change of variables

In integral calculus, sometimes directly integrating a function in terms of its original variable can be complex, which is why a change of variables is employed. This involves substituting one variable with another to simplify the integration process.

• In the provided solution, \( u = \sin x \) was used to substitute for \( \sin x \), while \( du = \cos x \, dx \) was the derivative.
• This change of variables shifted the integral from an expression involving trigonometric functions to a basic algebraic form \( \int \frac{1}{u^2 + 2u} \, du \).

The change of variables helps express the integral in a simpler or more familiar form. It allows us to utilize algebraic techniques or employ integral tables effectively, thus streamlining the resolution of otherwise daunting integral problems.